In signal processing, a filter is a device or process that removes some unwanted components or features from a signal. Filtering is a class of signal processing, the defining feature of filters being the complete or partial suppression of some aspect of the signal. Most often, this means removing some frequencies or frequency bands. However, filters do not exclusively act in the frequency domain; especially in the field of image processing many other targets for filtering exist.
Explanation:
Answer:
The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
Explanation:
The distance that the truck starts slowing down = 80 ft from the stop sign
Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.
u = initial velocity of the truck = 40 mph = 58.667 ft/s
v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)
x = horizontal distance covered during the deceleration
a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration
v² = u² + 2ax
0² = 58.667² + 2(-12)(x)
24x = 3441.816889
x = 143.41 ft
143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?
The truck will not be able to stop in time.
==> First lets convert all variables to SI units
1 mph = 0.45m/s
40mph = 40 miles per hour = 40 x 0.45 m/s
40mph = 18m/s
1 ft = 0.3048m
80 ft = 80 x 0.3048m
80 ft = 24.38m
Also;
12ft/s² = 12 x 0.3048m/s²
12ft/s² = 3.66m/s²
==> Now, consider one of the equations of motion as follows;
v² = u² + 2as -----------------(i)
Where;
v = final velocity of motion
u = initial velocity of motion
a= acceleration/deceleration of motion
s = distance covered during motion
Using this equation, lets calculate the distance, s, covered during the acceleration;
We know that;
v = 0 [since the truck comes to a stop]
u = 40mph = 18m/s
a = -12ft/s² = -3.66m/s² [the negative sign shows that the truck decelerates]
Substitute these values into equation (i) as follows;
0² = 18² + 2 (-3.66)s
0 = 324 - 7.32s
7.32s = 324
s =
s = 44.26m
The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m). This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.
B. Either vertically or horizontally polarized antennas may be used for transmission or reception
C. FM voice is unusable
D. Both the transmitting and receiving antennas must be of the same polarization
Answer:
B
Explanation:
The fact that skip signals refracted from the ionosphere are elliptically polarized is a result of either vertically or horizontally polarized antennas may be used for transmission or reception.
Answer:
Safety check is defined as rounding to make sure that the patients and the milieu (patients living quarters) is secured and free of harmful items that can be used to hurt someone.
Answer and Explanation:
The computation is shown below
Total average cost + total variable cost = total cost
Let number of output be x
So,
Total fixed average cost = x × $30
Total variable cost = x × $15
Total cost = $2,500
Therefore,
$20 × x + $30 × x = $2,500
50 × x = $2,500
x = 50
Now the total variable cost is
= 50 × $20
= $1,000
And, the fixed cost is
= 50 × $30
= $1,500
Answer:
12 4
Explanation:
because the production average is variable
Answer:
The problem is that the pumps would consume more energy than the generators would produce.
Explanation:
Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.
A pump uses electricity to add energy to the water to send it to a higher potential energy state.
Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.
What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.